People's Education Press will learn the first volume of ninth grade mathematics, and Hebei Education Press will learn the twenty-ninth chapter of ninth grade mathematics.

Definition: An integral equation has an unknown number, and the highest order of the unknown number is 2. Such an equation is called an unary quadratic equation.

The transformation from a linear equation to a quadratic equation is a qualitative change. Usually, quadratic equation is much more complicated in concept and solution than linear equation.

General form: ax 2+bx+c = 0 (a ≠ 0)

There are four general solutions:

Formula method (direct Kaiping method)

4. Matching method

3. Formula method

Factorial decomposition method

5. Cross multiplication

Cross multiplication can decompose some quadratic trinomials. The key of this method is to decompose the quadratic coefficient A into the product of two factors a 1 and A2 A 1. A2, decompose the constant term c into two factors, the product of c 1 and C2? C2, and make a 1c2+a2c 1 just a linear term b, then it can be directly written as a result: when decomposing factors in this way, we should pay attention to observation and try to realize that its essence is the inverse process of binomial multiplication. When the first coefficient is not 1, it often needs to be tested many times, so be sure to pay attention to the sign of each coefficient.

example

Example 1 Factorization 2x 2-7x+3.

Analysis: first decompose the quadratic coefficient and write it in the upper left corner and lower left corner of the crosshair, then decompose the constant term and divide it into two parts.

Don't write it in the upper right corner and lower right corner of the crosshair, and then cross multiply to find the algebraic sum to make it equal to the coefficient of the first term.

Quadratic coefficient decomposition (positive factor only):

2= 1×2=2× 1;

Decomposition of constant term:

3= 1×3=3× 1=(-3)×(- 1)=(- 1)×(-3).

Draw a cross line to represent the following four situations:

1 1

╳

2 3

1×3+2× 1

=5

1 3

╳

2 1

1× 1+2×3

=7

1 - 1

╳

2 -3

1×(-3)+2×(- 1)

=-5

1 -3

╳

2 - 1

1×(- 1)+2×(-3)

=-7

After observation, the fourth case is correct, because after cross multiplication, the algebraic sum of the two terms is exactly equal to the coefficient of the first term -7.

Solution 2x 2-7x+3 = (x-3) (2x- 1).

Generally speaking, for the quadratic trinomial ax2+bx+c(a≠0), if the quadratic term coefficient A can be decomposed into the product of two factors, that is, a=a 1a2, the constant term C can be decomposed into the product of two factors, that is, c=c 1c2, and A/KLOC-.

a 1 c 1

? ╳

a2 c2

a 1c2+a2c 1

Cross-multiply diagonally, and then add to get a 1c2+a2c 1. If it is exactly equal to the first term coefficient b of the quadratic trinomial ax2+bx+c, that is, a 1c2+a2c 1=b, the quadratic trinomial can be decomposed into two factors a65438+.

ax2+bx+c =(a 1x+c 1)(a2x+C2)。

The way to help us decompose the quadratic trinomial by drawing cross lines like this is usually called cross multiplication.

Example 2 Factorizing 6x 2-7x-5.

Analysis: According to the method of example 1, the quadratic term coefficient 6 and the constant term -5 are decomposed and arranged respectively, and there are eight different arrangement methods, one of which is

2 1

╳

3 -5

2×(-5)+3× 1=-7

Is correct, so the original polynomial can be factorized by cross multiplication.

Solution 6x 2-7x-5 = (2x+ 1) (3x-5)

It is pointed out that through the examples of 1 and 2, it can be seen that when a quadratic trinomial factor whose quadratic coefficient is not 1 is solved by cross integration, it often needs many observations to determine whether the factor can be solved by cross integration.

For the quadratic trinomial with quadratic coefficient of 1, cross multiplication can also be used to decompose the factors. At this time, you only need to consider how to decompose the constant term. For example, if x 2+2x-15 is decomposed, the cross multiplication is

1 -3

╳

1 5

1×5+ 1×(-3)=2

So x 2+2x- 15 = (x-3) (x+5).

Example 3 Factorization 5x 2+6xy-8y 2.

Analysis: This polynomial can be regarded as a quadratic trinomial about x, and-8y 2 is regarded as a constant term. When we decompose the coefficients of the quadratic term and the constant term, we only need to decompose 5 and -8, and then use the crosshairs to decompose them. After observation, we chose a suitable group, namely

1 2

? ╳

5 -4

1×(-4)+5×2=6

Solution 5x 2+6xy-8y 2 = (x+2y) (5x-4y).

It is pointed out that the original formula is decomposed into two linear formulas about x and y.

Example 4 Factorization (x-y)(2x-2y-3)-2.

Analysis: This polynomial is the product of two factors and the difference of another factor. Only by multiplying the polynomials first can the deformed polynomials be factorized.

Q: What are the characteristics of the factorial of the product of two products? What is the simplest method of polynomial multiplication?

A: If the common factor 2 is proposed for the first two items in the second factor, it will become 2(x-y), which is twice that of the first factor. Then multiply (x-y) as a whole, the original polynomial can be transformed into a quadratic trinomial about (x-y), and the factor can be decomposed by cross multiplication.

Solution (x-y)(2x-2y-3)-2

=(x-y)[2(x-y)-3]-2

=2(x-y) ^2-3(x-y)-2

=[(x-y)-2][2(x-y)+ 1]

=(x-y-2)(2x-2y+ 1)。

1 -2

╳

2 1

1× 1+2×(-2)=-3

It is pointed out that decomposing (x-y) into a whole is another application of holistic thinking method in mathematics.

Example 5 x 2+2x- 15

Analysis: Constant term (-15)

(-5) or (-3)(5), in which only the sum of -3 and 5 in (-3)(5) is 2.

=(x-3)(x+5)

Summary: ① factorization of x2+(p+q) x+pq formula.

The characteristics of this kind of quadratic trinomial formula are: the coefficient of quadratic term is1; Constant term is the product of two numbers; The coefficient of a linear term is the sum of two factors of a constant term. So we can directly decompose some quadratic trinomial factors with the coefficient of1:x 2+(p+q) x+pq = (x+p) (x+q).

② Factorization of KX2+MX+N formula

If it can be decomposed into k = AC, n = BD and AD+BC = M, then

kx^2+mx+n=(ax+b)(cx+d)

A b

╳

c d

1, direct Kaiping method:

The direct Kaiping method is a method to solve a quadratic equation with a direct square root. Solving (x-m)2=n (n≥0) by direct Kaiping method.

The solution is an equation of x = m.

Example 1. Solve the equation (1) (3x+1) 2 = 7 (2) 9x2-24x+16 =1.

Analysis: (1) This equation is obviously easy to do by direct flattening, (2) The left side of the equation is completely flat (3x-4)2, and the right side =11>; 0, so

This equation can also be solved by direct Kaiping method.

(1) solution: (3x+ 1)2=7×

∴(3x+ 1)2=5

∴ 3x+ 1 = (be careful not to lose the solution)

∴x=

The solution of the original equation is x 1=, x2=.

(2) Solution: 9x2-24x+16 =11.

∴(3x-4)2= 1 1

∴3x-4=

∴x=

The solution of the original equation is x 1=, x2=.

2. Matching method: solve the equation ax2+bx+c=0 (a≠0) by matching method.

First, move the constant c to the right of the equation: AX2+BX =-C.

Convert the quadratic term to 1: x2+x =-

Add half the square of the first order coefficient on both sides of the equation: x2+x+( )2=- +( )2.

The left side of the equation becomes completely flat: (x+ )2=

When b2-4ac≥0, x+=

∴x= (this is the root formula)

Example 2. Solving Equation 3x2-4x-2=0 by Matching Method

Solution: Move the constant term to the right of equation 3x2-4x=2.

Transform the quadratic term into 1: x2-x =

Add half the square of the coefficient of the first order term on both sides of the equation: x2-x+( )2= +( )2.

Formula: (x-)2=

Direct square: x-=

∴x=

The solution of the original equation is x 1=, x2=.

3. Formula method: convert the quadratic equation of one variable into a general form, and then calculate the value of the discriminant △=b2-4ac. B2-4ac≥0, release all items.

Substitute the values of coefficients A, B and C into the formula x=(b2-4ac≥0) to get the root of the equation.

Example 3. Solving Equation 2x2-8x=-5 by Formula Method

Solution: Change the equation into a general form: 2x2-8x+5=0.

∴a=2，b=-8，c=5

B2-4ac =(-8)2-4×2×5 = 64-40 = 24 & gt； 0

∴x= = =

The solution of the original equation is x 1=, x2=.

4. Factorial decomposition method: the equation is deformed into a form with one side zero, and the quadratic trinomial on the other side is decomposed into the product of two linear factors, so that,

Two linear factors are equal to zero respectively, and two linear equations are obtained. The roots obtained by solving these two linear equations are two of the original equations.

Root. This method of solving a quadratic equation with one variable is called factorization.

Example 4. Solve the following equation by factorization:

( 1)(x+3)(x-6)=-8(2)2 x2+3x = 0

(3) 6x2+5x-50=0 (optional study) (4)x2-2(+)x+4=0 (optional study)

(1) Solution: (x+3)(x-6)=-8 Simplified sorting.

X2-3x- 10=0 (the equation has a quadratic trinomial on the left and zero on the right).

(x-5)(x+2)=0 (factorization factor on the left side of the equation)

∴x-5=0 or x+2=0 (converted into two linear equations)

∴x 1=5,x2=-2 is the solution of the original equation.

(2) Solution: 2x2+3x=0

X(2x+3)=0 (factorize the left side of the equation by increasing the common factor)

∴x=0 or 2x+3=0 (converted into two linear equations)

∴x 1=0, x2=- is the solution of the original equation.

Note: Some students easily lose the solution of x=0 when doing this kind of problem. It should be remembered that there are two solutions to the quadratic equation of one variable.

(3) Solution: 6x2+5x-50=0

(2x-5)(3x+ 10)=0 (pay special attention to symbols when factorizing by cross multiplication).

2x-5 = 0 or 3x+ 10=0.

∴x 1=, x2=- is the solution of the original equation.

(4) Solution: x2-2(+ )x+4 =0 (∵4 can be decomposed into 2.2, ∴ this problem can be factorized).

(x-2)(x-2 )=0

∴x 1=2, x2=2 is the solution of the original equation.

5. Cross multiplication

The formula in the form of y = x 2+(p+q) x+pq can be factorized.

The characteristics of this kind of quadratic trinomial formula are: the coefficient of quadratic term is1; Constant term is the product of two numbers; The coefficient of a linear term is the sum of two factors of a constant term. So we can directly decompose some quadratic trinomial factors with the coefficient of1:x 2+(p+q) x+pq = (x+p) (x+q).

Binary Quadratic Equation: An integral equation with two unknowns, the highest degree of which is 2.

[Edit this paragraph] Note

Generally speaking, a linear equation with n unknowns is an equation with unknown terms of 1, and the coefficient of linear terms is not equal to 0.

N-dimensional linear equations are composed of several N-dimensional linear equations (except one-dimensional linear equations);

One-dimensional A-degree equation is an equation with an unknown term whose highest degree is A (except one-dimensional linear equation);

One-dimensional A-degree equation is an equation composed of several one-dimensional A-degree equations (except one-dimensional linear equation);

N-dimensional A-degree equation is an equation with n unknowns and the highest degree of the unknowns is A (except one-dimensional linear equation);

N-dimensional A-degree equation is an equation composed of several N-dimensional A-degree equations (except one-dimensional linear equation);

Among equations (groups), equations (groups) with more unknowns than equations are called indefinite equations (groups), and such equations (groups) generally have countless solutions.

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